PROP. 34. THEOR. B The opposite sides and angles of a parallelogram are equal to one another, and the diagonal bisects it. Given ABCD, a parallelogram, A i.e., given AB || DC, and joined. BC, 2D = _ B, LDAB = _ DCB, and A ADC A ABC. :: AB is || DC, and AC meets them, ...BAC = = LACD (1. 29); :: AD is || BC, and AC meets them, :: LDAC = L ACB (I. 29); and AC is common to triangles ADC, ABC; :: AADC = A ABC (I. 26), :: the diagonal bisects the parallelogram; and :. also AB DC, AD BC, and 2D = 2B; and . DAB - DCB, .: L BAC and 2 DAC are respectively equal to L ACD and L ACB. Q.E.D. 99. The diagonals of a parallelogram bisect each other. 100. The line joining the middle points of two sides of a triangle is parallel to the base, and equal to half the base of the triangle. 101. The lines joining the middle points of the adjacent sides of a quadrilateral form a parallelogram. 102. If a straight line be drawn parallel to the base of a triangle, bisecting one side, it shall also bisect the other side. 103. The diagonals of a square or a rhombus bisect each other at right angles. 104. If the opposite sides or opposite angles of a quadrilateral are equal, the figure is a parallelogram. PROP. 37. THEOR. triangles on the base parallels BC, AD. To prove A ABC A DBC. B Produce AD both ways; through B draw BE || AC (1. 31), and through C draw CF || BD. Then EBCA, DBCF are parallelograms on the same base BC, and between the same parallels BC and EF. :: IEC = O BF (I. 35). A ABC is half | EC (1. 34), and A DBC is half | BF, :: A ABC = ADBC (Ax. 7). Q.E.D. the PROP. 38. THEOR. H same F To mye A ABC = A DEF. D both ways; through B draw BG CA 1 through F draw FH" ED. I are parallelograms, and they are equal uther (1. 36): half I GC (1. 34), and A DEF is half | EH, AABC = ADEF Ar. TQ.E.D. PROP. 39. THEOR. A B Join EC. : AABC and A EBC are on the same base BC, and between the same parallels BC and AE; :: A ABC = A EBC (I. 37). But A ABC = ADBC (given), , : EBC = ADBC, which is impossible. :: AE is not || BC, and no other line through A but AD can be parallel to BC. :. AD is || BC. Q.E.D. 112. If a straight line is drawn from the vertex of a triangle to bisect the base, the triangle is divided into two parts equal in area. 113. In I. 39, show that AE must meet BD or BD produced. 114. If from two points in a straight line equal lines are drawn on the same side of it, making equal angles with the given line, the line which joins their other extremities is parallel to the given line. 115. If of the four triangles into which the diagonals divide a quadrilateral, two, whose bases are opposite sides of the quadrilateral, are equal, then the other two sides of the quadrilateral are parallel. 116. Two equal triangles are on the same base and on opposite sides of it; prove that the line joining their vertices is bisected by the base or base produced. 117. Equal triangles between the same parallels are on equal bases. For Euclid I. 30, see Appendix. PROP. 30. THEOR. Straight lines which are parallel to the same straight line are parallel to one another. Given AB || CD, and EF || B CD. с meet; then there will be :: AB is || EF. Q.E.D. E PROP. 31. PROB. Through a given point to draw a straight line parallel to a given straight line. Given st. line AB and pt. C. F To draw through C a st. line || AB. Take any point D in AB, D and join CD. At C in CD make the – ECD = _ CDB (I. 23). Produce EC. : _ECD LCDB, and these are alternate angles, :: EF is || AB (1. 27), and has been drawn through the given point C. Q.E.F. 73. Prove Prop. 30 by drawing a line to intersect the given lines. 74. If a straight line be parallel to two straight lines which meet at a point, these two shall be in the same straight line. PROP. 32. THEOR. If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles, and the three interior angles of every triangle are together equal to two right angles. Given A ABC having A + LB. B L ACB Through C draw CE || AB (1. 31). :: AC meets the parallels AB, CE, :: LA LACE (I. 29). : BD meets the parallels AB, CE, :: – ECD = LB (I. 29). :: LACD LA + LB. To each of these equals add – ACB. :: LACD + L ACB LA + 2B + - ACB. But L ACD + L ACB 2 Ls. (I. 13). :: LA + LB + L ACB = 2 L.s. Q.E.D. 75. A straight line parallel to the base of an isosceles iangle and intersecting the sides, or the sides produced, forms with them an isosceles triangle. 76. Through a given point to draw a st. line to make with a given st. line an angle equal to a given angle. How many such lines are there? 77. If a straight line be drawn through the vertex of an isosceles triangle parallel to the base, it shall bisect the exterior angle at the vertex. 78. In a right-angled triangle find a point in the hypotenuse AB so that its distance from AC equals its distance from B. 79. In an isosceles triangle find two points in the equal sides equidistant from the extremities of the base and from one another. |